Question

A woman pulls a 7.87 kg suitcase, initially at rest, with a 29.2 N force along the handle at 62.7 degrees. Friction pulls back at 11.7 N, and the suitcase moves 8.44 m. What is the total work done? PLEASE HELP

Answer 1

Work done (WD) is - Net Force 'F' times the displacement 's'.
\[WD = \text{F . s}\]
So, here you need to find the component of the force acting in the direction of teh displacement.

f = frictional force
F = Force which is being used to pull the suitcase
Fx= Component of 'F' in 'x' direction.
Fy= Component of 'F' in 'y' direction.
W= Weight of suitcase.

Now you need to resolve the forces such that you have the *net force* in the direction of the displacement, that is the *x-direction*. This *net force in 'x'* multiplied by the displacement gives the work done.

Can you do this?